#pragma GCC optimize("Ofast,unroll-loops")
#include <bits/stdc++.h>
using namespace std;
int dsu[50005];
int sz[50005];
int K = 400;
int find(int n) {
if (dsu[n] == n) return n;
return dsu[n] = find(dsu[n]);
}
void merge(int a, int b) {
a = find(a);
b = find(b);
if (a != b) {
if (sz[a] < sz[b]) swap(a, b);
dsu[b] = a;
sz[a] += sz[b];
}
}
int dt[100005][3];
int N, M;
int res[100005];
bool seen[50005];
vector<int> adj[50005];
int ans[850];
int CURR = 0;
struct Query {
int op, a, b;
};
set<tuple<int, int, int>> edges;
void dfs(int n) {
CURR += sz[n];
seen[n] = 1;
for (int i:adj[n]) if (!seen[i]) dfs(i);
}
void solve(vector<Query> &batch) {
// find which edges are not impacted
vector<tuple<int, int, int>> todo;
vector<int> changed;
for (int i = 0; i < batch.size(); i ++) {
Query Q = batch[i];
if (Q.op == 1) changed.push_back(Q.a);
else todo.push_back({Q.b, Q.a, i});
}
sort(todo.begin(), todo.end());
for (int i:changed) {
edges.erase({dt[i][0], dt[i][1], dt[i][2]});
}
auto ptr = edges.begin();
for (int i = 1; i <= N; i ++) dsu[i] = i, sz[i] = 1;
// Calculate edges
memset(ans, 0, sizeof(ans));
for (auto [w, n, id]:todo) {
// add all the edges
while (ptr != edges.end()) {
auto [a, b, c] = *ptr;
if (a <= w) {
merge(b, c);
ptr ++;
} else {
break;
}
}
// find the extra merges
for (int i:changed) res[i] = dt[i][0];
for (int i = 0; i < id; i ++) {
Query Q = batch[i];
if (Q.op == 1) {
res[Q.a] = Q.b;
}
}
for (int i:changed) {
if (res[i] <= w) {
// make edge
int t1 = find(dt[i][1]), t2 = find(dt[i][2]);
adj[t1].push_back(t2);
adj[t2].push_back(t1);
}
res[i] = 0;
}
CURR = 0;
dfs(find(n));
ans[id] = CURR;
// clear the result
seen[find(n)] = 0;
for (int i:changed) {
int t1 = find(dt[i][1]), t2 = find(dt[i][2]);
adj[t1].clear();
adj[t2].clear();
seen[t1] = 0;
seen[t2] = 0;
}
}
for (int i:ans) if (i != 0) cout << i << "\n";
// push this batch of data over
for (Query Q:batch) if (Q.op == 1) dt[Q.a][0] = Q.b;
for (int i:changed) edges.insert({dt[i][0], dt[i][1], dt[i][2]});
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cin >> N >> M;
if (M == 100000) K = 820;
for (int i = 1; i <= M; i ++) {
cin >> dt[i][1] >> dt[i][2] >> dt[i][0];
dt[i][0] = -dt[i][0];
edges.insert({dt[i][0], dt[i][1], dt[i][2]});
}
vector<Query> batch;
int Q;
cin >> Q;
auto start = clock();
int bruh = 0, tot = 0;
for (int i = 1; i <= Q; i ++) {
int a, b, c;
cin >> a >> b >> c;
if (a == 1) bruh ++;
tot ++;
c = -c;
batch.push_back({a, b, c});
if (bruh == K or tot == 800) {
solve(batch);
batch.clear();
tot = 0, bruh = 0;
}
}
if (!batch.empty()) solve(batch);
}
